Home › CBSE Class 12 › Calculus › $\displaystyle\int \dfrac{1}{\sqrt{1 - x^2}}\,dx…
$\displaystyle\int \dfrac{1}{\sqrt{1 - x^2}}\,dx$ equals:
A$\log|1 - x^2| + C$
B$\sin^{-1} x + C$
C$\tan^{-1} x + C$
D$\cos^{-1} x + C$
Answer & Solution
Correct answer: B. $\sin^{-1} x + C$
$\int \dfrac{1}{\sqrt{a^2 - x^2}}\,dx = \sin^{-1}(x/a) + C$, which with $a = 1$ gives $\sin^{-1} x + C$. The integral of the negative integrand would give $\cos^{-1} x + C$, since $\dfrac{d}{dx}\sin^{-1} x = +\dfrac{1}{\sqrt{1-x^2}}$ and $\dfrac{d}{dx}\cos^{-1} x = -\dfrac{1}{\sqrt{1-x^2}}$.
Related questions
A tank is draining and a student is asked how fast the depth falls when the volume is chanThe notation used for a family of antiderivatives, complete with its constant, is the:Knowing an object's velocity, a student recovers its position by finding an:A function whose derivative is the given function is called its:Calculators and computers rely on that same tangent-based idea when they find:An iterative technique for finding zeroes, built on tangent line approximations, is:The rule that resolves such a limit by differentiating numerator and denominator separatelA limit produces the form zero over zero, so its behaviour cannot be read off directly. Th